Problem 97 - Chemistry and Chemical Reactivity

Question

Potassium perchlorate is prepared by the following sequence of reactions: $$\mathrm{Cl}_{2}(\mathrm{g})+2 \mathrm{KOH}(\mathrm{aq}) \rightarrow \mathrm{KCl}(\mathrm{aq})+\mathrm{KClO}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell)$$ $$\begin{aligned}3 \mathrm{KClO}(\mathrm{aq}) & \rightarrow 2 \mathrm{KCl}(\mathrm{aq})+\mathrm{KClO}_{3}(\mathrm{aq}) \\\4 \mathrm{KClO}_{3}(\mathrm{aq}) & \rightarrow 3 \mathrm{KClO}_{4}(\mathrm{aq})+\mathrm{KCl}(\mathrm{aq})\end{aligned}$$ What mass of \(\mathrm{Cl}_{2}(\mathrm{g})\) is required to produce \(234 \mathrm{kg}\) of \(\mathrm{KClO}_{4} ?\)

Step-by-Step Solution

Answer

Approximately 359 kg of \(\mathrm{Cl}_{2}\) is required.

Step 1: Understanding the chemical reactions

The preparation of potassium perchlorate involves three sequential chemical reactions. Our goal is to determine the mass of chlorine gas \(\mathrm{Cl}_{2}\) needed to produce a specific amount of potassium perchlorate \(\mathrm{KClO}_{4}\).

Step 2: Analyze the conversion of compounds

According to the reactions, 1 mole of \(\mathrm{Cl}_{2}\) eventually results in the formation of \(\frac{1}{3}\) moles of \(\mathrm{KClO}_{4}\). This relationship needs to be kept in mind for further calculations.

Step 3: Calculate the moles of \(\mathrm{KClO}_{4}\)

First, calculate the moles of \(\mathrm{KClO}_{4}\) with the given mass. The molar mass of \(\mathrm{KClO}_{4}\) is approximately 138.55 g/mol. Convert 234 kg of \(\mathrm{KClO}_{4}\) to grams and then to moles:\[m = \frac{234,000 \text{ g}}{138.55 \text{ g/mol}} \approx 1690 \text{ moles}\].

Step 4: Relate moles of \(\mathrm{KClO}_{4}\) to moles of \(\mathrm{Cl}_{2}\)

Using the ratio from Step 2, 1 mole of \(\mathrm{Cl}_{2}\) produces \(\frac{1}{3}\) mole of \(\mathrm{KClO}_{4}\). Therefore, \(1690\) moles of \(\mathrm{KClO}_{4}\) requires \(1690 \times 3 = 5070\) moles of \(\mathrm{Cl}_{2}\).

Step 5: Calculate the mass of \(\mathrm{Cl}_{2}\)

The molar mass of \(\mathrm{Cl}_{2}\) is approximately 70.90 g/mol. Calculate the mass of \(\mathrm{Cl}_{2}\) using the number of moles we determined:\[m = 5070 \text{ moles} \times 70.90 \text{ g/mol} \approx 359,163 \text{ g} = 359.163 \text{ kg}\].

Key Concepts

Chemical Reactions

Chemical reactions are integral to the preparation of potassium perchlorate. In this process, three distinct reactions are used in sequence to ensure the complete transformation of chlorine gas (\(\mathrm{Cl}_{2}\)) into potassium perchlorate (\(\mathrm{KClO}_{4}\)). Understanding these reactions helps clarify how each compound participates in the overall preparation process.

Firstly, chlorine reacts with potassium hydroxide (\(\mathrm{KOH}\)) to form potassium chloride (\(\mathrm{KCl}\)) and hypochlorite (\(\mathrm{KClO}\)). This initial step sets up the formation of more complex chlorine-containing compounds through intermediate stages. As the reactions proceed, the hypochlorite ions are eventually converted into potassium chlorate (\(\mathrm{KClO}_{3}\)) and finally into potassium perchlorate.

The transformations involve oxidation and reduction processes, where chlorine experiences changes in oxidation states. Understanding these sequential reactions is crucial for calculating the amount of reagents required and predicting the products formed. This sequence must be followed precisely to obtain the desired end product.

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. In the context of this exercise, stoichiometry helps us determine how much chlorine gas is needed to produce a specific amount of potassium perchlorate.

To begin, one must comprehend the stoichiometric coefficients from the balanced reactions. These coefficients indicate the ratio in which reactants combine and products form. Here, for every one mole of chlorine gas needed, a fractional mole of potassium perchlorate is ultimately produced. By establishing these relationships, known as mole ratios, calculations can be made to translate from moles of products to moles of reactants.

This exercise specifically involves converting moles of potassium perchlorate back to moles of chlorine gas by using the stoichiometric relationship given by the chemical equations. Such precision in calculation ensures the reactants are neither wastage nor deficit, allowing chemists to conduct their reactions more efficiently.

Molar Mass Calculations

Molar mass calculations are pivotal when converting between mass and moles of a compound. If you're dealing with chemical reactions, you often need to convert how much you have measured physically into a form that's easier to work with for chemical equations, which is where moles come in.

To calculate moles of potassium perchlorate (\(\mathrm{KClO}_{4}\)) from a given mass, first convert kilograms to grams because molar mass is expressed in grams per mole. With the known molar mass of potassium perchlorate, which is approximately 138.55 g/mol, divide the total mass of the compound by this number to find the number of moles.

For example, 234 kg of potassium perchlorate converts to 234,000 g when measured in grams. Dividing this by the molar mass gives you the number of moles. These calculations are crucial since they lay the groundwork for determining how reactants relate to each other in any given reaction. It's through understanding molar mass that chemists can accurately tailor amounts for reactions, optimizing both cost and safety.